Proof composition of analytic functions is analytic

Notice that

$$\displaystyle\lim_{z\to z_0}\dfrac{gf(z)-gf(z_0)}{z-z_0} =\displaystyle\lim_{z\to z_0} \left[\dfrac{gf(z)-gf(z_0)}{z-z_0} \times\dfrac{f(z)-f(z_0)}{f(z)-f(z_0)}\right] $$

Where $z_0\in D$

This is really a pointwise result; we don't need analyticity in full open sets.

Suppose $f'(a),g'(f(a))$ both exist. Then $(g\circ f)'(a) = g'(f(a))f'(a).$

Proof: First, $f$ is continuous at $a$ since $f'(a)$ exists. Second, there exists $r>0$ and $M$ such that

$$|g(w)-g(f(a))| \le M|w-f(a)|, w \in D(f(a),r).$$

Suppose $f'(a) = 0.$ Because $f$ is continuous at $a,$ $f(z) \in D(f(a),r)$ for $z$ near $a.$ For such $z,z\ne a,$ we have

$$\left| \frac{g(f(z)) - g(f(a))}{z-a}\right| \le M\left| \frac{f(z)-f(a)|}{z-a}\right|.$$

The right side $\to 0$ and we see the desired result holds.

If $f'(a)\ne 0,$ then $f(z) \ne f(a)$ for $z$ in some punctured disc centered at $a.$ This allows for the "fake proof" to actually work: For such $z,$

$$\frac{g(f(z)) - g(f(a))}{z-a}= \frac{g(f(z)) - g(f(a))}{f(z)-f(a)}\cdot \frac{f(z)-f(a)}{z-a}.$$

Now let $z\to a$ to bring it on home.